$ f_{n}(x)=\sqrt{nx}\arctan\frac{1}{nx}$. Is the convergence uniform for $x\gt0$?

Question

Set, for $E=x\gt0$, $\quad f_{n}(x)=\sqrt{nx}\arctan\frac{1}{nx}$, $\quad n=1,2,3, \dots$. Is the convergence uniform on $E$?

I have pieced together the following solution from another example in the the textbook.

As $\quad f=\lim_{n\to\infty}f_{n}(x)=0.$

And $\quad f_{n}(\frac{1}{n})=\sqrt{1}\arctan(1)=\frac{\pi}{4}$

Then $\quad ||f_{n}-f||_{E} \geq \frac{\pi}{4}$.

Hence $\quad \lim_{n\to\infty}||f_{n}-f||_{E}$ can not converge to zero thus $f_{n}$ is not uniformly convergent on $E$.

Assuming my solution is correct? My question is how can I be sure that the inequality $ ||f_{n}-f||_{E} \geq \frac{\pi}{4}$ is true?

(Sidenote - The $x=\frac{1}{n}$ is just an educated guess, as taking the derivative of $||f_{n}-f||$ got messy and I could not see an obvious value of $x$ that would give me a maximum.)

Answer 1

Yes, the sequence $(f_n)_n$ does not converge uniformly in $\mathbb{R}^+$ to the pointwise limit $f=0$. Note that, by letting $t=nx$ we have $x>0$ iff $t>0$ and $$\sup_{x>0}|f_n(x)-f(x)|=\sup_{x>0}\sqrt{nx}\arctan\frac{1}{nx}=\sup_{t>0}\sqrt{t}\arctan\frac{1}{t}\geq \sqrt{1}\arctan\frac{1}{1}=\frac{\pi}{4}.$$

Answer 2

You can be sure because$$(\forall n\in\mathbb{N}):f_n\left(\frac1n\right)=\arctan1=\frac\pi4$$and because you are right when you state that $f\equiv0$.